Question

Find the specified nth term in the expansion of the binomial. (Write the expansion in descending powers of x.)

Answer 1

(x+5y)^10, n=4

Answer 2

Are you familiar with this ?

http://en.wikipedia.org/wiki/Binomial_coefficient

Answer 3

yes, partially, i'm just learning about it now. i haven't tried that pascal triangle yet, but the problem wants me to use binomial expansions i think

Answer 4

Here is the definition of binomial expansion:

http://en.wikipedia.org/wiki/Binomial_expansion

Looks like you are supposed to use pascal's triangle - I'm not aware of another way, but satellite might :)

Answer 5

@fiddlearound i always get the nth term confused

Answer 6

the other way is to write out what the coefficients are

Answer 7

the 4th term here will look like
\[\text{some number}\times x^7y^3\]

Answer 8

and the question is, what is the "some number" you need
\[\dbinom{10}{3}\times 5^3\] the
\[5^3\] because you really have
\[(5y)^3\]

Answer 9

so your real job is to cube 5 and get 125, and also to find 10 choose 3. do you know how to find that?

Answer 10

it's 15,000 x^7y^3

Answer 11

ok but do you know how to get it?

Answer 12

thanks for helping out!

Answer 13

yw

Answer 14

yeah, i think i understand, i have to move on to the next problem in my school work though, i have a ton to get done

Answer 15

just for the sake of completeness
\[\dbinom{10}{3}=\frac{10\times 9\times 8}{3\times 2}=10\times 3\times 4=120\]

Answer 16

yep, in short:

\[\left(\begin{matrix}10 \\ 3\end{matrix}\right)\times x ^{10-3} (5y) ^{3}=\frac{10!}{3!*(10-3)!}*125*x^7y^3=120*125*x^{7}y^3=15000x^7y^3\]